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This is a follow-up question to Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? . My motivation is to construct a statistic whose rate convergence to a Gaussian will be very slow and as such explore types of convergence which are not encapsulated by the Berry-Esseen' theorem. We therefore define the following statistic: \begin{equation} S_n := \frac{\left(\sum\limits_{k=1}^n f^{-1}(X_k) - \frac{3}{2} e \cdot n\right)}{e\cdot \sqrt{n \log(\log(n))}} \end{equation} where $X_k$ are i.i.d. uniformly distributed in $(0,1)$ and the function $f()$ is defined as follows: \begin{equation} f(x) := \frac{e^2}{2} \cdot \frac{1+\log(x)}{x^2 \log(x)^2} 1_{x \ge e} \end{equation} Now, the probability density of $f(X)$ is as follows: \begin{eqnarray} \rho_{f^{-1}(X)}(z) &=& \int\limits_0^1 \delta(z - f^{-1}(x)) dx =-\int\limits_{e}^\infty \delta(z-u) f^{'}(u) du= -f^{'}(z) 1_{z \ge e}\\ &=& \frac{e^2}{2} \cdot \frac{2+3 \log(z)+2 \log(z)^2}{z^3 \log(z)^3}1_{z \ge e} \end{eqnarray} From this we readily get the moments: \begin{eqnarray} E\left[ f^{-1}(X) \right] = \frac{3}{2} e\\ E\left[ (f^{-1}(X))^2 \right] = \infty \end{eqnarray} We also get the characteristic function. It reads: \begin{eqnarray} \kappa_{f^{-1}(X)}(k) = e^{\imath k e}+ \imath k \frac{e}{2} e^{\imath k e}-k^2 \frac{e^2}{2}\cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta \end{eqnarray} for $0<k<1$.

Note: The last integral on the right hand side is for me hard to crack. However numerical computations suggest that: \begin{equation} \lim_{k\rightarrow 0} \frac{1}{\log(\log(1/k))} \cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta = 1 \end{equation} Indeed by using the integral representation of the Gamma function along with integration by parts we quickly establish the following identity: \begin{eqnarray} (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) = \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) + \sum\limits_{n=1}^\infty \frac{(\imath k)^n}{n!}\cdot \frac{e^{n-\delta}}{\delta-n} \end{eqnarray}

Now clearly \begin{eqnarray} &&\int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta =\\ && \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) \right) d\delta + O(k)\\ &&= \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} - \frac{(-\imath k)^\delta}{\delta} \right) d\delta + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \left.\left( Ei(-\delta) - Ei(-A \delta)\right)\right|_0^\infty+ \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \log(-A) + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k) \end{eqnarray} where $A=-\log(-\imath k)= \imath \pi/2 - \log(k)$. Now we have checked numericaly that the integral in the middle above decays monotonically when $k\rightarrow 0$. Since now $\log(-A) = \log(-\imath \pi/2 + \log(k))= \log(-\imath \pi/2-\log(1/k)) \rightarrow \log(-\log(1/k)) = -\imath \pi/2 + \log(\log(1/k)) \rightarrow \log(\log(1/k))$ when $k\rightarrow 0$ the claim is established.

Now we check that our test statistic is properly normalized.

Define $c_n:=\sqrt{n\log(\log(n))}$. Indeed we have: \begin{eqnarray} &&\log\left( \kappa_{S_n}(k)\right) =\\ && -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ \kappa_{f^{-1}(X)}(\frac{k}{e c_n})\right] \\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ e^{\imath \frac{k}{c_n}}(1+\imath \frac{k}{2 c_n}) - \frac{1}{2} \frac{k^2}{c_n^2} \log(1+\log(c_n)-\log(k))\right]\\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ 1+\imath \frac{3}{2} \frac{k}{c_n} - \frac{k^2}{c_n^2} + O(\frac{k^3}{c_n^3}) - \frac{1}{2} \frac{k^2}{c_n^2} \log(1+\log(c_n)-\log(k)) \right]\\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + \imath k \frac{3}{2} \frac{n}{c_n} + \left(\frac{1}{8} - \frac{1}{2} \log(1+\log(c_n)-\log(k))\right) \frac{k^2}{c_n^2} n + O(\frac{k^3}{c_n^3})\\ &&= \left(\frac{1}{8} - \frac{1}{2} \log(1+\log(c_n)-\log(k))\right) \frac{k^2}{c_n^2} n + O(\frac{k^3}{c_n^3}) \end{eqnarray} Now for the statistic to be properly normalized we have to have: \begin{equation} \lim_{n\rightarrow \infty} \frac{n}{c_n^2} \log(\log(c_n)) = 1 \end{equation} which is indeed the case as one can readily check by plugging the definition of $c_n$ into the lhs.

Now, I carried out a Monte Carlo simulation and computed the sample Cumulative Distribution Function (CDF) of our statistic and plotted it along with the CDF of a standardized Gaussian distribution with the former and the later being plotted in Blue and Purple respectively. Here I took $n=5,10,15$ and in each case I used $m=1000$ realizations. The figures are below:

CDFs at $n=5$

CDFs at $n=10$

CDFs at $n=15$

I have used the following Mathematica code to produce those figures:

m = 1000; n = 15; delta = 1/10;
bins = Table[-5 + delta/2 + j delta, {j, 1, (10 - delta)/delta}];
limD = CDF[NormalDistribution[0, 1], bins];
X = RandomReal[{0, 1}, {m, n}];
x =.; {t0, Y} = 
 Timing[(x /. 
    Map[First[
       NSolve[(E^2 (1 + Log[x]))/(2 x^2 Log[x]^2) == # && x > E, x, 
        Reals]] &, X, {2}])];
ll = (Total[#] & /@ Y - 3/2 E n)/(E Sqrt[n Log[Log[n]]];
emp = EmpiricalDistribution[ll];
DD = CDF[emp, bins];
pl = ListPlot[Transpose[{bins, #}] & /@ {DD, limD}, ImageSize -> 800, 
   LabelStyle -> {15, FontFamily -> "Arial"}, 
   BaseStyle -> {15, FontFamily -> "Bold"}, 
   PlotLabel -> "n=" <> ToString[n]];
Export["LimitBehavior1_n_" <> ToString[n] <> ".jpg", pl, "JPEG"];
Import["LimitBehavior1_n_" <> ToString[n] <> ".jpg"]

Having said all this my question is the following. What is the rate of convergence of our statistic towards a Gaussian. To be specific we are asking about the behavior of the supremum norm of the difference in CDFs for large values of $n$.

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  • $\begingroup$ The Berry-Esseen theorem's statement involves random variables who have finite first three moments. But it seems like your random variable doesn't even have a variance... If you mean for the $\log\log$ term in the denominator to be a correction for this, then the distributions the statistic $S_n$ works on now vary with $n$. How is this going to lead to a counterexample? $\endgroup$ – enthdegree Apr 5 '18 at 18:07
  • $\begingroup$ Both in my example and in the other example linked to this question the second moments do not exist. This is compensated by different normalization factor,i.e $\sqrt{n \log(\log(n))}$ in my case and by $\sqrt{n \log(n)}$ in the case of the linked example. Now the really interesting question is the convergence ratio since it will turn out to be very slow, even much slower then in the linked example where it was $O(\log(\log(n)))/\log(n))$. Another lesson is that simulations are almost useless for exploring such cases since one has to go to very large $n$ values to see anything. $\endgroup$ – Przemek Repetowicz Apr 5 '18 at 18:28
  • $\begingroup$ I agree that if it does converge then the rate will be much slower than Berry-Esseen, but it seems that you've just confirmed the penultimate sentence in my first comment: With this normalization factor, the source distributions the statistic works over vary with the statistic's index $n$. The Berry-Esseen theorem doesn't make a statement about this situation. $\endgroup$ – enthdegree Apr 5 '18 at 18:38
  • 1
    $\begingroup$ @enthdegree: Yes of course, the theorem in question does not describe this case and that is why I wanted to look into it to analyze it further.I am almost certain that there exist a whole family of functions $f()$ that lead to extremely slow convergence rates being related to iterated logarithms of $n$. $\endgroup$ – Przemek Repetowicz Apr 6 '18 at 9:33

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